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Choose the holomorphic square root in the complex upper half-plane whose boundary value is positive on $0<x<1$. Its primitive has derivative $F_1^{\prime }(z)=1/\sqrt{z(1-z^2)}$. The integral from the boundary basepoint zero is well-defined by a limiting path: the local singularity is only $z^{-1/2}$.

The three finite prevertices $-1,0,1$ have derivative exponent $-1/2$, hence image angle $\pi /2$. At infinity $F_1^{\prime }(z)=O(z^{-3/2})$, so the local coordinate $1/z$ also gives image angle $\pi /2$. These are the four corners of a Schwarz-Christoffel mapping to a rectangle. To establish that it is a square rather than a general rectangle, putThe length along $(-1,0)$ is also $A$ by $x\mapsto -x$. The length along $(1,\infty )$ isby $x=1/t$, and the fourth side has the same length by symmetry. With the chosen branch, the boundary derivatives on the intervals $(-\infty ,-1),(-1,0),(0,1),(1,\infty )$ have phases $-1,-i,1,i$, respectively. ThereforeThe extended real boundary maps continuously and once around this square. Each side is traversed monotonically, and the four corner limits are finite. The argument principle, applied after small indentations around the boundary singularities, counts one inverse image of each interior value and zero of each exterior value. Since $F_1^{\prime }$ never vanishes in the complex upper half-plane, those interior inverse images are simple. Thus

Take a real modulus $0<k<1$ and complementary modulus $k^{\prime }=\sqrt{1-k^2}$. The Jacobi elliptic sine is constructed by inverting a Schwarz-Christoffel mapping. The main reason for the Schwarz-Christoffel derivative is local angle behavior: if a prevertex $a_j$ corresponds to a polygon angle $\pi {\alpha }_j$, straightening that corner gives $U(w)-U(a_j)\sim C_j(w-a_j{)}^{{\alpha }_j}$. Hence $U^{\prime }$ has exponent ${\alpha }_j-1$. Along each straight boundary side its argument is constant. Dividing $U^{\prime }$ by ${\prod }_j(w-a_j{)}^{{\alpha }_j-1}$ removes those angle changes; Schwarz reflection extends the quotient across the real boundary and the corner singularities. With all vertices finite, its behavior at infinity is also regular, because the sum of the derivative exponents is $-2$. It is therefore a nonzero constant on the sphere. This yieldsThis argument explains the exponents, the branch choices and the role of boundary reflection, rather than just writing down the formula.

For a rectangle, four angles are $\pi /2$, so choose prevertices $-1/k,-1,1,1/k$ and normalize the inverse map asUse the branch positive on $(-1,1)$. SetThe two middle corner values are $U(\pm 1)=\pm K$. The right side length isas the substitution $t=(1-k^{\prime 2}{s}^2{)}^{-1/2}$ shows. The selected branch makes the increment along that side $i{K}^{\prime }$. Thus the other corner values are $K+i{K}^{\prime }$ and $-K+i{K}^{\prime }$. The two boundary intervals through infinity have combined horizontal length $2K$; the substitution $t=1/(ks)$ gives length $K$ from $1/k$ to infinity. In particular $U(\infty )=i{K}^{\prime }$. As in the square mapping, the boundary goes once around a rectangle, and the argument principle gives a conformal bijectionIts inverse $s(z)=\mathrm{sn}(z,k)$ maps that rectangle onto the complex upper half-plane, and near zero has $s(0)=0$, $s^{\prime }(0)=1$. Differentiating the inverse relation givesThese identities continue meromorphically.

All four rectangle edges have real images, so Schwarz reflection extends $s$ across them. Reflection in the bottom edge gives $s(\bar{z})=\overline{s(z)}$. Reflection in the right edge, combined with the first reflection, gives $s(2K-z)=s(z)$. The even inverse-integral derivative gives $s(-z)=-s(z)$, and henceReflection in the top edge and then the bottom gives $s(z+2i{K}^{\prime })=s(z)$. The resulting period lattice of the Jacobi elliptic sine isTo see the poles, use $U^{\prime }(w)\sim -1/(k{w}^2)$ near the boundary point infinity. Then $U(w)-i{K}^{\prime }\sim 1/(kw)$, so the inverse has a simple pole at $i{K}^{\prime }$ with residue $1/k$. The real half-period shift gives another pole at $2K+i{K}^{\prime }$ with opposite residue. Reflection tiles the plane with copies of the rectangle, so these are exactly two simple poles per displayed fundamental cell. They also show there is no additional period: a period must permute these two pole classes; exchanging them would shift by $2K$ modulo the displayed lattice, which reverses the function's sign and residues rather than preserving it.

For any finite $w$, integrate $s^{\prime }(z)/(s(z)-w)$ around a fundamental parallelogram avoiding its zeros and poles. The opposite edges cancel by periodicity, so the number of zeros of $s-w$ equals the number of poles, namely two. Therefore every value has two inverse images modulo periods, counted with multiplicity. The value infinity likewise has the two simple poles as preimages. Usually the finite-value preimages are distinct; the involution $z\mapsto 2K-z$ interchanges them. At $w=\pm 1,\pm 1/k$, a critical point supplies one double inverse image instead. For example $s(K)=1$, $s^{\prime }(K)=0$, $s^{\prime \prime }(K)=-(1-k^2)\ne 0$. Thus the printed “exactly two” requires the standard multiplicity convention; it would be false if interpreted as two distinct solutions at these branch values of a holomorphic map.

Now choose $z_0=K$ and put $f(z)=s(z+K)$. Since $s(2K-z)=s(z)$, $f$ is even, elliptic for $\Lambda$, and has degree two. The Weierstrass elliptic function for the same lattice is even and has one double pole per cell, hence also degree two. Its generic fibers are precisely $\{z,-z\}$. Consequently an even $f$ is constant on those fibers and descends to a meromorphic function $R$ of $\wp$. At the four fixed classes of $z\mapsto -z$, the even local Laurent expansion makes this descent meromorphic in the squared local coordinate. A meromorphic function on the Riemann sphere is rational, so $f=R(\wp )$. Generic map degrees multiply:Thus $R$ has degree one and the degree-two even elliptic function conclusion isThe determinant condition follows because $f$ is nonconstant. This proves the existence requested, with the explicit allowable shift $z_0=K$.

For a polygonal domain with finite real prevertices $a_j$ and interior angles $\pi {\alpha }_j$, the Schwarz-Christoffel formula has the formwith branches chosen in the complex upper half-plane. A prevertex at infinity is interpreted by the corresponding limiting version of the formula. The exponent records the angle because locally $F-F(a_j)$ behaves as $(z-a_j{)}^{{\alpha }_j}$.

For a slit from $0$ to $\tau$ in the upper half-plane, put $\vartheta =\arg \tau \in (0,\pi )$. The two approaches to its base are separate boundary vertices with angles $\pi -\vartheta$ and $\vartheta$, and the tip has angle $2\pi$. With ordered prevertices $a<c<b$ mapping respectively to the base, tip and base, the Schwarz-Christoffel mapping therefore hasThe tip's factor is a simple zero of the boundary derivative, not an interior critical point.

For the specified power product, the principal branches givethroughout the complex upper half-plane, so $g$ maps into that half-plane. Logarithmic differentiation givesThere are no zeros of this derivative in the upper half-plane, so the map is locally conformal there. Its boundary values are negative real for $x<-1$, positive real for $x>1$, andThe positive scalar factor increases from zero to its maximum at $c$ and then decreases to zero, since its logarithmic derivative is $-k/(1-x)+(1-k)/(1+x)$. Therefore the tip isThe real boundary runs from the negative real axis to zero, out to $\tau$, back to zero, and then along the positive real axis.

For completeness, these boundary facts give global injectivity and surjectivity. At infinity,Apply the argument principle on a large upper half-disc, indenting the real prevertices by small semicircles. For a fixed $w$ in the upper half-plane off $[0,\tau ]$, the limiting image contour is the real boundary with the slit traversed once in each direction, followed by a large upper semicircle. Its winding number about $w$ is one. The two slit traversals cancel, so $g(z)-w$ has exactly one zero in the upper half-plane. There are no poles.

Nor can an interior point map onto the open slit. Each point of that slit has two real preimages, one on either side of $c$, with nonzero boundary derivatives. Local continuation across the real line maps the upper half-neighborhoods at these preimages onto the two different sides of the slit. An additional interior preimage would persist for nearby off-slit points and contradict the already proved one-preimage count. The tip cannot be an interior image either: the open mapping theorem would then place nearby open-slit points in the interior image. Finally, $g$ never vanishes inside the half-plane. Thus $g$ is a conformal bijection onto $S(\tau )$.

This is the power-product map to a slit half-plane. Its derivative is exactly the Schwarz-Christoffel formula with $a=-1$, $b=1$, $c=1-2k$, $\vartheta =\pi k$ and $C=1$. The additive constant is fixed by the boundary value $g(1)=0$. Thus the power-product expression is an explicit antiderivative of the corresponding Schwarz-Christoffel mapping, with this normalization.

The finite corners have interior angle $3\pi /2$, hence exterior-angle parameters $-1/2$. The channel end at $Z=0$ is logarithmic. Their Schwarz-Christoffel mapping is consequentlyChoose $R(Z)=\sqrt{Z^2-1}$ analytic in the upper half-plane with $R(Z)\sim Z$ at upper infinity. In particular $R(iy)=i\sqrt{1+y^2}$ for $y>0$. A direct differentiation givesHence $f=A+CH$ with $H=R+{\tan }^{-1}(1/R)$.

The inverse trigonometric functions here is the branch obtained by analytic continuation from $Z>1$ through the upper half-plane, not an independent Cauchy principal value evaluation at each point. Its endpoint values give $H(1)=\pi /2$, $H(-1)=-\pi /2$. Imposing $f(1)=1$ and $f(-1)=-1$ therefore gives $A=0$, $C=2/\pi$:Along the positive imaginary axis, $H$ is purely imaginary and tends to $-i\infty$ as $Z\to 0$; at upper infinity $H\sim Z$. The real intervals $(0,1)$ and $(-1,0)$ trace the channel walls $x=1$ and $x=-1$ downward, while the outer intervals trace the horizontal boundary. This verifies the stated end assignments as well as the constants. The map is the Schwarz-Christoffel map of a half-plane joined to a strip.

A boundary point of the source half-plane that maps to a polygon vertex. If that corner has angle $\pi \alpha$, the Schwarz-Christoffel mapping has local behavior $F(z)-F(a)\sim C(z-a{)}^{\alpha }$ at its prevertex, so the derivative has exponent $\alpha -1$. Infinity may also be a prevertex, expressed in the local coordinate $1/z$.